# Probability and Statistics: In a certain small town, Comcast has 100000 subscribers.

#### tommy123456789

##### New member
I need help with these 3 problems:

1)In a certain small town, Comcast has 100000 subscribers. According to Census data, 40 percent of the subscribers are single parent households.
A simple random sample of 400 households were selected for a survey. The percentage of single parent households in the sample will be around ___40_______, give or take _________or so.
Estimate the chance that between 35 percent and 45 percent of the households in the sample are single parent households________
Ok I was able to figure out part 1 other 2 keep being told it is the wrong answer

2) A simple random sample of 400 persons is taken to estimate the percentage of women in the Boston Teacher's Union whose total membership is 8000.
Of those surveyed, 261 teachers are females. We can be 95 percent confident that between _________ percent and _________ percent of the teachers in the union are females.

3)A college has 25000 students. Based on student personal information data, approximately 55 percent are Democrats.

A simple random sample of 900 households were selected for a survey. The percentage of Democrats in the sample will be around _55____, give or take
_______or so.
Estimate the chance that between 52.45 percent and 57.55 percent of the students in the sample are Democrats. ____
any help appreciated!!!!!!!!!!:?:

#### HallsofIvy

##### Elite Member
Each of these problems has two possible outcomes (single parent or not, woman or not, Democrat or not) so they are binomial distributions. In each the numbers involved are large so these look like exercises in using the normal distribution approximation to the binomial distribution. A binomial distribution in which the two outcomes have probability p and q= 1- p, with large N, can be approximated by a normal distribution with mean Np and standard deviation $$\displaystyle \sqrt{npq}$$.

For example in problem 1 the probability a subscriber is a single parent family is p= 40%= 0.4 and the probability a subscriber is not a single parent family is q= 1- 0.4= 0.6. N= 100000 so this binomial distribution can be approximated by a normal distribution with mean number of single parent families is 0.4(100000)= 40000 and standard deviation $$\displaystyle \sqrt{0.4(0.6)(100000)}= \sqrt{24,000}= 155$$ approximately. Have you tried that?